Question

For an n-channel silicon MOSFET with 10 nm gate oxide thickness, the substrate sensitivity \( \left( \frac{\partial V_T}{\partial |V_{BS}|} \right) \) is found to be 50 mV/V at a substrate voltage \[ |V_{BS}| = 2 \, \text{V}, \text{ where } V_T \text{ is the threshold voltage of the MOSFET. Assume that } |V_{BS}| \gg 2 \Phi_B, \text{ where } q \Phi_B \text{ is the separation between the Fermi energy level } E_F \text{ and the intrinsic level } E_i \text{ in the bulk. Parameters given are:} \] \[ \text{Electron charge } (q) = 1.6 \times 10^{-19} \, \text{C}, \quad \text{Vacuum permittivity } (\varepsilon_0) = 8.85 \times 10^{-12} \, \text{F/m}, \] \[ \text{Relative permittivity of silicon } (\varepsilon_{si}) = 12, \quad \text{Relative permittivity of oxide } (\varepsilon_{ox}) = 4. \] \text{The doping concentration of the substrate is:}

• A. \( 7.37 \times 10^{15} \, \text{cm}^{-3} \)
• B. \( 4.37 \times 10^{15} \, \text{cm}^{-3} \)
• C. \( 2.37 \times 10^{15} \, \text{cm}^{-3} \)
• D. \( 9.37 \times 10^{15} \, \text{cm}^{-3} \)

Hint:

When calculating the doping concentration from substrate sensitivity, use the relationship between the threshold voltage sensitivity and the capacitance of the oxide.

Answer 1

The Correct Option is A

We are given the substrate sensitivity \( \left( \frac{\partial V_T}{\partial |V_{BS}|} \right) \) and the values for the electron charge, permittivities, and relative permittivities of silicon and oxide. To find the doping concentration of the substrate, we use the following relationship: \[ \frac{\partial V_T}{\partial |V_{BS}|} = \frac{q}{C_{ox}} \sqrt{2 q \varepsilon_{si} N_A}, \] where \( C_{ox} = \frac{\varepsilon_{ox}}{t_{ox}} \) is the oxide capacitance, and \( N_A \) is the doping concentration. Solving this equation with the given values, we find the doping concentration to be \( 7.37 \times 10^{15} \, \text{cm}^{-3} \). Final Answer: \[ \boxed{7.37 \times 10^{15} \, \text{cm}^{-3}}. \]